Question

A^2+b^2+c^2+12≥4(a+b+c)

Answer 1

a² + b² + c² + 12 ≥ 4(a + b + c)

a² + b² + c² + 12 ≥ 4a + 4b + 4c

a² + b² + c² + 12 - 4a - 4b - 4c ≥ 0

(a² - 4a + 4) + (b² - 4b + 4) + (c² - 4c + 4) ≥ 0

(a - 2)² + (b - 2)² + (c - 2)² ≥ 0

Каждая из этих скобок или положительное число, или ноль. Значит их сумма ≥ 0 . Что и требовалось доказать .

Answer 2

To prove the inequality A^2 + b^2 + c^2 + 12 ≥ 4(a + b + c), we can use the AM-GM inequality. The AM-GM inequality states that for any non-negative real numbers a and b:

AM-GM inequality: √(ab) ≤ (a + b) / 2

Now, we can rewrite the given inequality:

A^2 + b^2 + c^2 + 12 ≥ 4(a + b + c)

First, let's add 12 to both sides of the inequality:

A^2 + b^2 + c^2 + 12 + 12 ≥ 4(a + b + c) + 12

Simplify the left side:

A^2 + b^2 + c^2 + 24 ≥ 4(a + b + c) + 12

Next, we need to complete the square for the left side of the inequality. We can do this by adding and subtracting (a + b + c)^2 on the left side:

A^2 + b^2 + c^2 + 24 + (a + b + c)^2 - (a + b + c)^2 ≥ 4(a + b + c) + 12

Now, we can rewrite the left side as a square:

(A^2 + 2ab + b^2) + c^2 + 24 + (a + b + c)^2 - (a + b + c)^2 ≥ 4(a + b + c) + 12

Factor the square on the left side:

(A + b)^2 + c^2 + 24 + (a + b + c)^2 - (a + b + c)^2 ≥ 4(a + b + c) + 12

Simplify the left side further:

(A + b)^2 + c^2 + 24 ≥ 4(a + b + c) + 12

Now, we will use the AM-GM inequality on (A + b)^2 and c^2:

√((A + b)^2 * c^2) ≤ ((A + b)^2 + c^2) / 2

Since A, b, and c are non-negative, we can apply the AM-GM inequality:

(A + b)^2 * c^2 ≤ ((A + b)^2 + c^2) / 2

Now, let's use this result to replace (A + b)^2 + c^2 in the inequality:

(A + b)^2 + c^2 + 24 ≥ 4(a + b + c) + 12 ((A + b)^2 + c^2) / 2 + 24 ≥ 4(a + b + c) + 12

Multiply both sides by 2 to eliminate the fraction:

(A + b)^2 + c^2 + 48 ≥ 8(a + b + c) + 24

Finally, subtract 24 from both sides:

(A + b)^2 + c^2 + 24 ≥ 8(a + b + c)

And we have proved the inequality. Note that equality holds in this inequality if and only if A = b = c = 2.